Alcohol Phenol And Ethers

This is a long answer type question as classified in NCERT Exemplar

Commercially, ethanol (C₂H₅OH) is made by fermenting carbohydrates, which is the earliest process. In the presence of an enzyme called invertase, sugar in molasses, sugarcane, or fruits like grapes is transformed to glucose and fructose (both of which have the formula C₆H₁₂O₆).

In the presence of another enzyme, zymase, present in yeast, glucose and fructose are fermented.

Grapes are used to make wine because they contain sugars and yeast. The amount of sugar in grapes increases as they ripen, and yeast forms on the outer peel. When grapes are crushed, sugar and enzymes collide, causing fermentation to begin. Fermentation takes place in the absence of oxygen in anaerobic circumstances. During fermentation, carbon dioxide is emitted.

This is a long answer type question as classified in NCERT Exemplar

11.84

The driving force of all the reactions given to the question is that the alkoxy group is an ortho and para directing group because it exerts its +R effect in the benzene ring. Para position being comparatively more stable than the ortho position is usually preferred because ortho position leads to stearic hindrance, hence the major product is mostly the para- substituted compound.

As seen from the resonating structures above the structure in which the negative charge is in the para position will form a more stable product when attacked by an electrophile. Hence in the following reactions, we will be considering that resonating structure as the starting material.

The mechanism is given below:

The driving force of all the reactions given to the question is that the alkoxy group is an ortho and para directing group because it exerts its +R effect in the benzene ring. Para position being comparatively more stable than the ortho position is usually preferred because ortho position leads to stearic hindrance, hence the major product is mostly the para-substituted

As seen from the resonating structures above the structure in which the negative charge is in the para position will form a more stable product when attacked by an electrophile. Hence in the following reactions, we will be considering that resonating structure as the starting material.

Nitration of anisole gives p-nitroanisole as the major product.

The mechanism is given below:

The driving force of all the reactions given to the question is that the alkoxy group is an ortho and para directing group because it exerts its +R effect in the benzene ring. Para position being comparatively more stable than the ortho position is usually preferred because ortho position leads to stearic hindrance, hence the major product is mostly the para-substituted

The driving force of all the reactions given to the question is that the alkoxy group is an ortho

As seen from the resonating structures above the structure in which the negative charge is in the para position will form a more stable product when attacked by an electrophile. Hence in the following reactions, we will be considering that resonating structure as the starting material. and para directing group because it exerts its +R effect in the benzene ring. Para position being comparatively more stable than the ortho position is usually preferred because ortho position leads to stearic hindrance, hence the major product is mostly the para-substituted compound.

The Friedel-Craft's acetylation of anisole gives 4- methoxyacetophenone as the major product.

11.83 

The reaction of HI with methoxymethane yields two different sets of products depending upon the initial amount of HI taken.

When equal moles of HI and methoxymethane are taken, a mixture of methyl alcohol and methyl iodide is

The mechanism is given below:

In the first step, methoxymethane reacts with hydrogen iodide to extract a proton to give the dimethyloxonium ion.

In the second step of the reaction, the Dimethyloxonium ion reacts with the iodide ion present to yield methyl iodide and methyl alcohol as the product via SN² pathway.

If an excess of HI is used the methyl alcohol formed in Step II is also converted into methyl iodide by the following mechanism : -

In the first step, methoxymethane reacts with hydrogen iodide to extract a proton to give the dimethyloxonium ion.

In the second step of the reaction the Dimethyloxonium ion reacts with the iodide ion present to yield methyl iodide and methyl alcohol as the product via SN²  pathway.

In the third step of the reaction Methyl alcohol formed above reacts with hydrogen iodide to extract a proton to give the protonated methyl alcohol which finally reacts in the fourth step with the iodide ion via SN² pathway to give methyl iodide and water as the product.